php - Cron Jobs 在运行时滞后于我的网站

Question

<分区>

Answer 1

第一个建议是将准备调用移出循环。更新将锁定您的表一段时间，因此仅在完成操作后更新...缓存数组中需要更新的内容并启动事务，执行所有更新而不是提交。 更新了邮件脚本的代码示例

<?php
require('functions.php');

global $mysqli;

$select = $mysqli->prepare("SELECT `email`, `username`, `id` FROM `users` WHERE `email` IS NOT NULL AND `email` <> '' AND `credits` < `credits_offered` AND `emailed_credits` = 0");
$select->execute();
$select->bind_result($email, $username, $id);

$users = array();

while($select->fetch())
{
    $users[] = array("id" => $id, "email" => $email, "username" => $username);
}

$u = array();

foreach($users as $user)
{
    $to      = $user['email'];
    $subject = '';
    $message = "";
    $headers = 'From: email@domain.com' . "\r\n" .
        'Reply-To: reply@domain.com';

    mail($to, $subject, $message, $headers);
    $u[] = $user['id'];

}
$update = $mysqli->prepare("UPDATE `users` SET `emailed` = 1 WHERE `id` = ?");
$update->bind_param('i', &$i);
$mysqli->autocommit(false);
foreach($u as $i)
    $update->execute();
$update->close();
$mysqli->commit();

哦，差点忘了，大多数 mysql 存储引擎都非常愚蠢，确保你使用 InnoDB，它有事务支持。或者，如果您使用“愚蠢的”存储引擎，如果您的 users.id 是 int，您仍然可以通过执行

来欺骗它进行单个事务来更新多行

$update = $mysqli->prepare('UPDATE users SET emailed=1 WHERE id IN (' . implode(',', $u) . ')');
$update->execute();
$update->close();

在关闭 foreach($users as $user) 语句之后